Question: What is the product of all the coordinates of all the points of intersection of the two circles defined by $x^2-2x +y^2-10y+25=0$ and $x^2-8x+y^2-10y+37=0$?
Explanation: Add $(-2/2)^2$ and $(-10/2)^2$ to the first equation and $(-8/2)^2$ and $(-10/2)^2$ to the second equation to find that the given equations are equivalent to \begin{align*}
(x^2-2x+1)+(y^2-10y+25)&=1\text{, and} \\
(x^2-8x+16)+(y^2-10y+25)&=4
\end{align*} which are equivalent to \begin{align*}
(x-1)^2+(y-5)^2 &=1^2, \\
(x-4)^2+(y-5)^2 &=2^2,
\end{align*} respectively.  Hence, the two circles have centers $(1,5)$ and $(4,5)$ respectively and radii $1$ and $2$ respectively. Since the centers of the circles are $3$ units apart and the sum of their radii is $3$, the two circles intersect at only one point.  We can see that $(2,5)$ is the desired intersection point, so our product is $2 \cdot 5 =\boxed{10}$.